Complex logarithm question

**The question**

Find the cartesian form of the principle value of:

$\displaystyle ((\frac{1 + i\sqrt{3}}{2})^{-3})^{1 - i}$

**My attempt**

$\displaystyle e^{(1 - i)Log(e^{(\frac{\pi i}{3})}^{-3})$

$\displaystyle e^{(1 - i)Log(e^{-\pi i}})$

$\displaystyle e^{(1 - i)(ln|1| + i(-\pi))}$ //uh oh, this is on a branch cut!

How do I proceed, if the principle Log is not defined on the negative real axis?

Thank you.

Re: Complex logarithm question

Not sure there's a problem here. The only place Log isn't defined is at $\displaystyle z=0$.

Re: Complex logarithm question

Quote:

Originally Posted by

**Glitch** **The question**

Find the cartesian form of the principle value of:

$\displaystyle ((\frac{1 + i\sqrt{3}}{2})^{-3})^{1 - i}$

**My attempt**

$\displaystyle e^{(1 - i)Log(e^{(\frac{\pi i}{3})}^{-3})$

$\displaystyle e^{(1 - i)Log(e^{-\pi i}})$

$\displaystyle e^{(1 - i)(ln|1| + i(-\pi))}$ //uh oh, this is on a branch cut!

How do I proceed, if the principle Log is not defined on the negative real axis?

Thank you.

$\displaystyle \displaystyle \left[\left(\frac{1 + i\sqrt{3}}{2}\right)^{-3}\right]^{1 - i} = \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i}$

$\displaystyle \displaystyle \left|\frac{1 + i\sqrt{3}}{2}\right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$

$\displaystyle \displaystyle \arg{\left(\frac{1 + i\sqrt{3}}{2}\right)} = \arctan{\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2 }}\right)} = \arctan{\left(\sqrt{3}\right)} = \frac{\pi}{3}$

$\displaystyle \displaystyle \begin{align*} \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i} &= \left(e^{\frac{\pi i}{3}}\right)^{-3 + 3i} \\ &= e^{(-1 + i)\pi i} \\ &= e^{-\pi - \pi i} \\ &= e^{-\pi}\cos{(-\pi)} + i\,e^{-\pi}\sin{(-\pi)} \\ &= -e^{-\pi} \end{align*}$

Re: Complex logarithm question

Thanks, that's the answer I eventually got. However the book answer is $\displaystyle -e^{\pi}$. I think they just added $\displaystyle 2\pi$ to the principle argument (not sure why).

Re: Complex logarithm question

Quote:

Originally Posted by

**Glitch** Thanks, that's the answer I eventually got. However the book answer is $\displaystyle -e^{\pi}$. I think they just added $\displaystyle 2\pi$ to the principle argument (not sure why).

It's because the principal argument is in the domain $\displaystyle \displaystyle (-\pi, \pi]$.

Re: Complex logarithm question

Quote:

Originally Posted by

**Glitch** However the book answer is $\displaystyle -e^{\pi}$. I think they just added $\displaystyle 2\pi$ to the principle argument (not sure why).

I think the answer to that must be that they wanted you to evaluate the expression in the order indicated by the bracketing:

$\displaystyle \Bigl[\Bigl(\frac{1 + i\sqrt{3}}{2}\Bigr)^{-3}\Bigr]^{1 - i} = \left[(e^{i\pi/3})^{-3}\right]^{1-i} = (e^{-i\pi})^{1-i}.$

At that stage, they want you to replace $\displaystyle e^{-i\pi}$ by its principal value $\displaystyle e^{i\pi}$ before raising it to the power $\displaystyle 1-i.$ Then you get $\displaystyle (e^{i\pi})^{1-i} = e^{(i+1)\pi} = -e^\pi$.

Re: Complex logarithm question

Quote:

Originally Posted by

**Prove It** $\displaystyle \displaystyle \left[\left(\frac{1 + i\sqrt{3}}{2}\right)^{-3}\right]^{1 - i} = \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i}$

$\displaystyle \displaystyle \left|\frac{1 + i\sqrt{3}}{2}\right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$

$\displaystyle \displaystyle \arg{\left(\frac{1 + i\sqrt{3}}{2}\right)} = \arctan{\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2 }}\right)} = \arctan{\left(\sqrt{3}\right)} = \frac{\pi}{3}$

$\displaystyle \displaystyle \begin{align*} \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i} &= \left(e^{\frac{\pi i}{3}}\right)^{-3 + 3i} \\ &= e^{(-1 + i)\pi i} \\ &= e^{-\pi - \pi i} \\ &= e^{-\pi}\cos{(-\pi)} + i\,e^{-\pi}\sin{(-\pi)} \\ &= -e^{-\pi} \end{align*}$

You got to be careful here; $\displaystyle ({z^\alpha})^\beta \ne z^{\alpha \beta}$ in general.

Re: Complex logarithm question

Quote:

Originally Posted by

**Glitch** Thanks, that's the answer I eventually got. However the book answer is $\displaystyle -e^{\pi}$. I think they just added $\displaystyle 2\pi$ to the principle argument (not sure why).

You can't just add stuff. Check what principal argument range the book is using.